MATH-181 (Calculus I) Curriculum

Session 04 - Infinite Limits and Limits at Infinity

Handout Worksheet
Fall Semester 2026
Session Objectives & Overview

By the end of this session, you will be able to:

  • Evaluate one-sided and two-sided infinite limits analytically using Directional Zero Sign Analysis and identify vertical asymptotes versus removable discontinuities.
  • Evaluate limits at infinity for polynomials and rational/radical functions using Global Growth Dominance filtering and determine horizontal asymptotes.
  • Resolve advanced end-behavior forms at infinity, including ∞−∞\infty - \infty∞−∞ indeterminate radical forms, negative infinity radical sign subtleties (x2=∣x∣=−x\sqrt{x^2} = |x| = -xx2​=∣x∣=−x), and asymptotic exponential saturation models.

Micro-Lecture

1. Infinite Limits & Physical Singularities

  • Mathematical Foundation: When direct substitution into a simplified quotient produces c0\frac{c}{0}0c​ (c≠0c \neq 0c=0), the function grows without bound (±∞\pm\infty±∞). The vertical line x=ax = ax=a is a Vertical Asymptote, representing an Infinite Discontinuity. We determine the direction (+∞+\infty+∞ vs −∞-\infty−∞) using directional signed zero analysis (0+0^+0+ vs 0−0^-0−).
  • Engineering Context: Infinite limits characterize localized physical singularities, such as mechanical resonance in civil infrastructure (unbounded vibration amplitudes when external driving frequencies match structural natural frequencies) or electrical dielectric breakdown and thermal runaway in high-voltage insulators.

2. Limits at Infinity & Global Growth Dominance

  • Mathematical Foundation: A Limit at Infinity (lim⁡x→±∞f(x)\lim_{x \to \pm\infty} f(x)limx→±∞​f(x)) analyzes the global end behavior of a function. By the principle of Growth Dominance, higher-power leading terms completely dominate lower-order terms as ∣x∣→∞|x| \to \infty∣x∣→∞. When this limit converges to a finite value LLL, the line y=Ly = Ly=L is a Horizontal Asymptote.
  • Engineering Context: Limits at infinity define physical ceilings, saturation capacities, and steady-state thresholds, including the maximum product throughput (Vmax⁡V_{\max}Vmax​) in chemical bioreactors and asymptotic response latencies in distributed computer cloud networks.

3. Advanced Asymptotic Forms: Radical Conjugates & Exponential Ceilings

  • Mathematical Foundation: Forms like ∞−∞\infty - \infty∞−∞ at infinity are indeterminate and require radical conjugate multiplication to convert into determinate ratios. When evaluating limits as x→−∞x \to -\inftyx→−∞ containing square roots, the absolute value rule x2=∣x∣=−x\sqrt{x^2} = |x| = -xx2​=∣x∣=−x introduces a critical negative sign.
  • Engineering Context: These methods allow aerospace engineers to compute the steady-state terminal velocity of re-entry vehicles under aerodynamic drag and enable computer engineers to program calibration offset margins for autonomous vehicle radar tracking.

Skill Block 1

Worked Example A: Graphical Limit and Asymptote Identification

Consider the graph of a piecewise rational function f(x)f(x)f(x) exhibiting asymptotes and boundary transitions:

  • As x→−4−x \to -4^-x→−4−, the curve shoots upwards without bound.
  • As x→−4+x \to -4^+x→−4+, the curve approaches an open circle at (−4,6)(-4, 6)(−4,6).
  • As x→3−x \to 3^-x→3−, the curve shoots upwards without bound.
  • As x→3+x \to 3^+x→3+, the curve shoots downwards without bound.
  • As x→∞x \to \inftyx→∞, the curve flattens along y=−2y = -2y=−2.
  • As x→−∞x \to -\inftyx→−∞, the curve flattens along y=1y = 1y=1.

(a) Evaluate lim⁡x→−4−f(x)\lim_{x \to -4^-} f(x)limx→−4−​f(x), lim⁡x→−4+f(x)\lim_{x \to -4^+} f(x)limx→−4+​f(x), lim⁡x→3−f(x)\lim_{x \to 3^-} f(x)limx→3−​f(x), lim⁡x→3+f(x)\lim_{x \to 3^+} f(x)limx→3+​f(x), lim⁡x→∞f(x)\lim_{x \to \infty} f(x)limx→∞​f(x), and lim⁡x→−∞f(x)\lim_{x \to -\infty} f(x)limx→−∞​f(x).
(b) Identify all vertical asymptotes.
(c) Identify all horizontal asymptotes.

Worked Solution:
  1. Part (a) — Directional and End-Behavior Limits:
    • lim⁡x→−4−f(x)=∞\lim_{x \to -4^-} f(x) = \inftylimx→−4−​f(x)=∞ (upward unbounded curve)
    • lim⁡x→−4+f(x)=6\lim_{x \to -4^+} f(x) = 6limx→−4+​f(x)=6 (approaches open hole coordinate)
    • lim⁡x→3−f(x)=∞\lim_{x \to 3^-} f(x) = \inftylimx→3−​f(x)=∞ (upward unbounded curve)
    • lim⁡x→3+f(x)=−∞\lim_{x \to 3^+} f(x) = -\inftylimx→3+​f(x)=−∞ (downward unbounded curve)
    • lim⁡x→∞f(x)=−2\lim_{x \to \infty} f(x) = -2limx→∞​f(x)=−2 (levels off along horizontal line)
    • lim⁡x→−∞f(x)=1\lim_{x \to -\infty} f(x) = 1limx→−∞​f(x)=1 (levels off along horizontal line)
  2. Part (b) — Vertical Asymptotes:
    • At x=−4x = -4x=−4, the left-hand limit is ∞\infty∞, so x=−4x = -4x=−4 is a vertical asymptote.
    • At x=3x = 3x=3, both one-sided limits are infinite (∞\infty∞ and −∞-\infty−∞), so x=3x = 3x=3 is a vertical asymptote.
  3. Part (c) — Horizontal Asymptotes:
    • As x→−∞x \to -\inftyx→−∞, f(x)→1  ⟹  y=1f(x) \to 1 \implies \mathbf{y = 1}f(x)→1⟹y=1.
    • As x→∞x \to \inftyx→∞, f(x)→−2  ⟹  y=−2f(x) \to -2 \implies \mathbf{y = -2}f(x)→−2⟹y=−2.

Worked Example B: Analytical Asymptote and Discontinuity Classification

Using limits, find all vertical and horizontal asymptotes for:

f(x)=2x23x−x2f(x) = \frac{2x^2}{3x - x^2}f(x)=3x−x22x2​

Worked Solution:
  1. Factor and Identify Candidate Singularities:
    f(x)=2x2x(3−x)f(x) = \frac{2x^2}{x(3 - x)}f(x)=x(3−x)2x2​
    The denominator is zero at x=0x = 0x=0 and x=3x = 3x=3.
  2. Test Candidate x=0x = 0x=0: Direct substitution gives 00\frac{0}{0}00​. Canceling the common factor xxx for x≠0x \neq 0x=0:
    lim⁡x→02x3−x=2(0)3−0=03=0\lim_{x \to 0} \frac{2x}{3 - x} = \frac{2(0)}{3 - 0} = \frac{0}{3} = 0x→0lim​3−x2x​=3−02(0)​=30​=0
    Because the limit is finite (000), x=0x = 0x=0 is a removable discontinuity (hole), not a vertical asymptote.
  3. Test Candidate x=3x = 3x=3: Substituting into the simplified form 2x3−x\frac{2x}{3 - x}3−x2x​:
    lim⁡x→32x3−x→60\lim_{x \to 3} \frac{2x}{3 - x} \to \frac{6}{0}x→3lim​3−x2x​→06​
    Because this is c0\frac{c}{0}0c​ (c≠0c \neq 0c=0), x=3x = 3x=3 is a vertical asymptote.
    • Left-hand limit (x→3−x \to 3^-x→3−): denominator 3−x→0+  ⟹  lim⁡x→3−f(x)=+60+=+∞3 - x \to 0^+ \implies \lim_{x \to 3^-} f(x) = \frac{+6}{0^+} = +\infty3−x→0+⟹limx→3−​f(x)=0++6​=+∞.
    • Right-hand limit (x→3+x \to 3^+x→3+): denominator 3−x→0−  ⟹  lim⁡x→3+f(x)=+60−=−∞3 - x \to 0^- \implies \lim_{x \to 3^+} f(x) = \frac{+6}{0^-} = -\infty3−x→0−⟹limx→3+​f(x)=0−+6​=−∞.
  4. Horizontal Asymptote Verification: Divide numerator and denominator by highest denominator power x2x^2x2:
    lim⁡x→±∞2x23x−x2=lim⁡x→±∞23x−1=20−1=−2\lim_{x \to \pm\infty} \frac{2x^2}{3x - x^2} = \lim_{x \to \pm\infty} \frac{2}{\frac{3}{x} - 1} = \frac{2}{0 - 1} = -2x→±∞lim​3x−x22x2​=x→±∞lim​x3​−12​=0−12​=−2
    • Vertical Asymptote: x=3x = 3x=3
    • Horizontal Asymptote: y=−2y = -2y=−2

Practice Problem 1: Concept & Reading Check

Open this Skill Block on the Memex to locate the information needed to answer the following:

  • (a) Under the Conceptual Framework, what is the fundamental mathematical distinction between direct substitution yielding c0\frac{c}{0}0c​ (c≠0c \neq 0c=0) versus yielding 00\frac{0}{0}00​?
  • (b) In Worked Example B, why is x=0x = 0x=0 classified as a removable discontinuity rather than a vertical asymptote, while x=3x = 3x=3 is classified as a vertical asymptote?
  • (c) Review the linked Engineering Applications section. How does a civil engineer use directional zero sign analysis to evaluate structural vibration amplitude in the Structural Resonance Threshold (Civil Engineering) model?
Solution Key:
  • (a) A non-zero numerator over an infinitesimally small denominator (c0\frac{c}{0}0c​) grows without bound (±∞\pm\infty±∞), defining a vertical asymptote. A 00\frac{0}{0}00​ form indicates that the numerator and denominator share a root, meaning the discontinuity is removable once the common factor is factored and canceled.
  • (b) Factoring f(x)=2x2x(3−x)=2x3−xf(x) = \frac{2x^2}{x(3-x)} = \frac{2x}{3-x}f(x)=x(3−x)2x2​=3−x2x​ (for x≠0x \neq 0x=0) reveals that lim⁡x→02x3−x=0\lim_{x \to 0} \frac{2x}{3-x} = 0limx→0​3−x2x​=0. Since the limit exists and is finite, x=0x = 0x=0 is a removable hole. At x=3x = 3x=3, substitution gives 60\frac{6}{0}06​, which diverges to ±∞\pm\infty±∞, confirming a vertical asymptote.
  • (c) In structural engineering, approaching the resonance frequency ω0=8 rad/s\omega_0 = 8\text{ rad/s}ω0​=8 rad/s from below yields ω<8  ⟹  ω2<64  ⟹  64−ω2>0\omega < 8 \implies \omega^2 < 64 \implies 64 - \omega^2 > 0ω<8⟹ω2<64⟹64−ω2>0. The signed zero approach gives 150+=+∞\frac{15}{0^+} = +\infty0+15​=+∞, proving vibration amplitude grows infinitely without damping.
Show your steps and calculations here.

Practice Problem 2

Evaluate the directional limit analytically:

lim⁡x→2+−5x−2\lim_{x \to 2^{+}} \frac{-5}{x-2}x→2+lim​x−2−5​

Solution Key:
  • Numerator: −5-5−5 (negative constant).
  • Denominator: As x→2+x \to 2^+x→2+, x>2x > 2x>2, so x−2>0x - 2 > 0x−2>0 (approaches 0+0^+0+).
  • Sign division: lim⁡x→2+−5x−2=−50+=−∞\lim_{x \to 2^+} \frac{-5}{x-2} = \frac{-5}{0^+} = -\inftylimx→2+​x−2−5​=0+−5​=−∞.
Show your steps and calculations here.

Practice Problem 3

Evaluate the directional limit analytically:

lim⁡x→3+43−x\lim_{x \to 3^{+}} \frac{4}{3-x}x→3+lim​3−x4​

Solution Key:
  • Numerator: +4+4+4 (positive constant).
  • Denominator: As x→3+x \to 3^+x→3+, x>3x > 3x>3, so 3−x<03 - x < 03−x<0 (approaches 0−0^-0−).
  • Sign division: lim⁡x→3+43−x=+40−=−∞\lim_{x \to 3^+} \frac{4}{3-x} = \frac{+4}{0^-} = -\inftylimx→3+​3−x4​=0−+4​=−∞.
Show your steps and calculations here.

Practice Problem 4

Evaluate the directional limit analytically:

lim⁡x→−4+3xx+4\lim_{x \to -4^{+}} \frac{3x}{x+4}x→−4+lim​x+43x​

Solution Key:
  • Numerator: As x→−4+x \to -4^+x→−4+, 3x→3(−4)=−12<03x \to 3(-4) = -12 < 03x→3(−4)=−12<0.
  • Denominator: As x→−4+x \to -4^+x→−4+, x>−4x > -4x>−4, so x+4>0x + 4 > 0x+4>0 (approaches 0+0^+0+).
  • Sign division: lim⁡x→−4+3xx+4=−120+=−∞\lim_{x \to -4^+} \frac{3x}{x+4} = \frac{-12}{0^+} = -\inftylimx→−4+​x+43x​=0+−12​=−∞.
Show your steps and calculations here.

Practice Problem 5

Evaluate the limit analytically:

lim⁡x→0+1x+5\lim_{x \to 0^{+}} \frac{1}{x+5}x→0+lim​x+51​

Solution Key:
  • Denominator: As x→0+x \to 0^+x→0+, x+5→5≠0x + 5 \to 5 \neq 0x+5→5=0.
  • Direct substitution is determinate: lim⁡x→0+1x+5=10+5=15\lim_{x \to 0^+} \frac{1}{x+5} = \frac{1}{0+5} = \frac{1}{5}limx→0+​x+51​=0+51​=51​.
Show your steps and calculations here.

Practice Problem 6

Evaluate the limit analytically:

lim⁡x→−5+(x+5)22x2+10x\lim_{x \to -5^{+}} \frac{(x+5)^{2}}{2x^{2}+10x}x→−5+lim​2x2+10x(x+5)2​

Solution Key:
  • Direct substitution gives 00\frac{0}{0}00​. Factor the denominator: 2x2+10x=2x(x+5)2x^2 + 10x = 2x(x + 5)2x2+10x=2x(x+5).
  • Simplify for x≠−5x \neq -5x=−5:
    (x+5)22x(x+5)=x+52x\frac{(x+5)^2}{2x(x+5)} = \frac{x+5}{2x}2x(x+5)(x+5)2​=2xx+5​
  • Re-evaluate the limit by direct substitution:
    lim⁡x→−5+x+52x=−5+52(−5)=0−10=0\lim_{x \to -5^+} \frac{x+5}{2x} = \frac{-5 + 5}{2(-5)} = \frac{0}{-10} = 0x→−5+lim​2xx+5​=2(−5)−5+5​=−100​=0
Show your steps and calculations here.

Practice Problem 7

Evaluate the trigonometric limit analytically:

lim⁡θ→π−csc⁡(θ)\lim_{\theta \to \pi^{-}} \csc(\theta)θ→π−lim​csc(θ)

Solution Key:
  • Rewrite using the reciprocal identity: csc⁡(θ)=1sin⁡(θ)\csc(\theta) = \frac{1}{\sin(\theta)}csc(θ)=sin(θ)1​.
  • As θ→π−\theta \to \pi^-θ→π−, θ\thetaθ is in Quadrant II (θ<π\theta < \piθ<π), where sin⁡(θ)>0\sin(\theta) > 0sin(θ)>0.
  • As θ→π\theta \to \piθ→π, sin⁡(θ)→0+\sin(\theta) \to 0^+sin(θ)→0+.
  • Sign division: lim⁡θ→π−1sin⁡(θ)=10+=∞\lim_{\theta \to \pi^-} \frac{1}{\sin(\theta)} = \frac{1}{0^+} = \inftylimθ→π−​sin(θ)1​=0+1​=∞.
Show your steps and calculations here.

Regroup 1

  • Key Reflection: Why does c0\frac{c}{0}0c​ (c≠0c \neq 0c=0) indicate a vertical asymptote, while 00\frac{0}{0}00​ requires further algebraic work before classifying the discontinuity?
  • Directional Sign Check: Emphasize that evaluating directional limits at vertical asymptotes never requires plugging in long decimal tables—identifying whether the denominator is 0+0^+0+ or 0−0^-0− and checking the numerator sign gives the exact sign of infinity immediately.

Skill Block 2

Worked Example A: Balanced Degree Rational Limit and Horizontal Asymptote

Evaluate the limit and state the equation of any horizontal asymptote:

lim⁡x→∞−5x3−2x+33x3+3x2−5x\lim_{x \to \infty} \frac{-5x^{3}-2x+3}{3x^{3}+3x^{2}-5x}x→∞lim​3x3+3x2−5x−5x3−2x+3​

Worked Solution:
  1. Identify Denominator Dominant Power: The highest power of xxx in the denominator is x3x^3x3.
  2. Divide Numerator and Denominator by x3x^3x3:
    lim⁡x→∞−5x3x3−2xx3+3x33x3x3+3x2x3−5xx3=lim⁡x→∞−5−2x2+3x33+3x−5x2\lim_{x \to \infty} \frac{\frac{-5x^3}{x^3} - \frac{2x}{x^3} + \frac{3}{x^3}}{\frac{3x^3}{x^3} + \frac{3x^2}{x^3} - \frac{5x}{x^3}} = \lim_{x \to \infty} \frac{-5 - \frac{2}{x^2} + \frac{3}{x^3}}{3 + \frac{3}{x} - \frac{5}{x^2}}x→∞lim​x33x3​+x33x2​−x35x​x3−5x3​−x32x​+x33​​=x→∞lim​3+x3​−x25​−5−x22​+x33​​
  3. Apply the Reciprocal Limit Rule (lim⁡x→∞cxp=0\lim_{x \to \infty} \frac{c}{x^p} = 0limx→∞​xpc​=0):
    −5−0+03+0−0=−53\frac{-5 - 0 + 0}{3 + 0 - 0} = -\frac{5}{3}3+0−0−5−0+0​=−35​
  4. State Horizontal Asymptote:
    • Limit: −53-\frac{5}{3}−35​
    • Horizontal Asymptote: y=−53y = -\frac{5}{3}y=−35​

Worked Example B: Fractional and Radical Power Growth Dominance

Evaluate the limit and state any horizontal asymptote:

lim⁡x→∞x3+41−5x\lim_{x \to \infty} \frac{\sqrt[3]{x}+4}{1-5\sqrt{x}}x→∞lim​1−5x​3x​+4​

Worked Solution:
  1. Convert Radicals to Fractional Exponents:
    x3=x1/3andx=x1/2\sqrt[3]{x} = x^{1/3} \quad \text{and} \quad \sqrt{x} = x^{1/2}3x​=x1/3andx​=x1/2
    The highest power of xxx in the denominator is x1/2x^{1/2}x1/2.
  2. Divide Numerator and Denominator by x1/2x^{1/2}x1/2:
    lim⁡x→∞x1/3x1/2+4x1/21x1/2−5x1/2x1/2=lim⁡x→∞x1/3−1/2+4x1/21x1/2−5=lim⁡x→∞1x1/6+4x1/21x1/2−5\lim_{x \to \infty} \frac{\frac{x^{1/3}}{x^{1/2}} + \frac{4}{x^{1/2}}}{\frac{1}{x^{1/2}} - \frac{5x^{1/2}}{x^{1/2}}} = \lim_{x \to \infty} \frac{x^{1/3 - 1/2} + \frac{4}{x^{1/2}}}{\frac{1}{x^{1/2}} - 5} = \lim_{x \to \infty} \frac{\frac{1}{x^{1/6}} + \frac{4}{x^{1/2}}}{\frac{1}{x^{1/2}} - 5}x→∞lim​x1/21​−x1/25x1/2​x1/2x1/3​+x1/24​​=x→∞lim​x1/21​−5x1/3−1/2+x1/24​​=x→∞lim​x1/21​−5x1/61​+x1/24​​
  3. Evaluate Limits of Reciprocal Terms: Since 1/6>01/6 > 01/6>0 and 1/2>01/2 > 01/2>0, both terms in the numerator vanish to 000:
    0+00−5=0−5=0\frac{0 + 0}{0 - 5} = \frac{0}{-5} = 00−50+0​=−50​=0
  4. State Horizontal Asymptote:
    • Limit: 000
    • Horizontal Asymptote: y=0y = 0y=0

Practice Problem 8: Concept & Reading Check

Open this Skill Block on the Memex to locate the information needed to answer the following:

  • (a) In the Conceptual Framework, what are the three outcomes in the rational degree comparison taxonomy, and what does each outcome imply for horizontal asymptotes?
  • (b) In Worked Example B, how were the radical expressions rewritten and compared to determine which power of xxx dominates in the denominator?
  • (c) Review the linked Engineering Applications section. How does growth dominance explain the maximum production speed ceiling Vmax⁡V_{\max}Vmax​ in the Industrial Bioreactor Output Bottlenecks (Chemical Engineering) model?
Solution Key:
  • (a) When the denominator degree is strictly larger (n<mn < mn<m), the fraction is bottom-heavy and shrinks to 000 (y=0y = 0y=0). When degrees are equal (n=mn = mn=m), growth rates are balanced and the limit is the ratio of leading coefficients (y=an/bmy = a_n/b_my=an​/bm​). When the numerator degree is strictly larger (n>mn > mn>m), the top grows faster and diverges to ±∞\pm\infty±∞ (no horizontal asymptote).
  • (b) By rewriting radicals as powers of xxx (x1/3x^{1/3}x1/3 vs x1/2x^{1/2}x1/2), the student identifies x1/2x^{1/2}x1/2 as the highest power in the denominator. Dividing all terms by x1/2x^{1/2}x1/2 results in positive powers in the denominator of each numerator term (x1/2−1/3=x1/6x^{1/2 - 1/3} = x^{1/6}x1/2−1/3=x1/6), driving the overall limit to 000.
  • (c) In enzyme saturation kinetics, the model V([S])=450[S]18+[S]V([S]) = \frac{450[S]}{18 + [S]}V([S])=18+[S]450[S]​ has balanced degrees in [S][S][S]. For very large substrate additions, lower-order constants (181818) become negligible compared to [S][S][S], producing a horizontal asymptote at Vmax⁡=450 kg/hV_{\max} = 450\text{ kg/h}Vmax​=450 kg/h.
Show your steps and calculations here.

Practice Problem 9

Evaluate the polynomial limit at negative infinity analytically:

lim⁡x→−∞(7x7−5x3+4x−1)\lim_{x \to -\infty} (7x^{7}-5x^{3}+4x-1)x→−∞lim​(7x7−5x3+4x−1)

Solution Key:
  • The leading term is 7x77x^77x7.
  • As x→−∞x \to -\inftyx→−∞, x7→(−∞)7=−∞x^7 \to (-\infty)^7 = -\inftyx7→(−∞)7=−∞ (an odd power preserves the negative sign).
  • Multiplying by +7+7+7: lim⁡x→−∞7x7=−∞\lim_{x \to -\infty} 7x^7 = -\inftylimx→−∞​7x7=−∞.
Show your steps and calculations here.

Practice Problem 10

Evaluate the polynomial limit at negative infinity analytically:

lim⁡x→−∞(1−2x+5x2−17x3−4x7)\lim_{x \to -\infty} (1-2x+5x^{2}-17x^{3}-4x^{7})x→−∞lim​(1−2x+5x2−17x3−4x7)

Solution Key:
  • The leading term (highest power) is −4x7-4x^7−4x7.
  • As x→−∞x \to -\inftyx→−∞, x7→−∞x^7 \to -\inftyx7→−∞.
  • Multiplying by negative coefficient −4-4−4: lim⁡x→−∞(−4x7)=(−4)(−∞)=∞\lim_{x \to -\infty} (-4x^7) = (-4)(-\infty) = \inftylimx→−∞​(−4x7)=(−4)(−∞)=∞.
Show your steps and calculations here.

Practice Problem 11

Evaluate the limit and state any horizontal asymptote (if none, state that):

lim⁡x→∞4x5+6x2−25−4x+2x3\lim_{x \to \infty} \frac{4x^{5}+6x^{2}-2}{5-4x+2x^{3}}x→∞lim​5−4x+2x34x5+6x2−2​

Solution Key:
  1. Divide by Highest Denominator Power (x3x^3x3):
    lim⁡x→∞4x5x3+6x2x3−2x35x3−4xx3+2x3x3=lim⁡x→∞4x2+6x−2x35x3−4xx2+2\lim_{x \to \infty} \frac{\frac{4x^5}{x^3} + \frac{6x^2}{x^3} - \frac{2}{x^3}}{\frac{5}{x^3} - \frac{4x}{x^3} + \frac{2x^3}{x^3}} = \lim_{x \to \infty} \frac{4x^2 + \frac{6}{x} - \frac{2}{x^3}}{\frac{5}{x^3} - \frac{4x}{x^2} + 2}x→∞lim​x35​−x34x​+x32x3​x34x5​+x36x2​−x32​​=x→∞lim​x35​−x24x​+24x2+x6​−x32​​
  2. Evaluate Component Limits:
    • Denominator approaches 0−0+2=20 - 0 + 2 = 20−0+2=2.
    • Numerator is dominated by 4x2→∞4x^2 \to \infty4x2→∞.
      ∞2=∞\frac{\infty}{2} = \infty2∞​=∞
  3. Conclusion:
    • Limit: ∞\infty∞
    • Horizontal Asymptote: None (function is top-heavy and grows without bound).
Show your steps and calculations here.

Practice Problem 12

Evaluate the rational limit and state the horizontal asymptote:

lim⁡x→∞8x4−3x2+12x4+7x−5\lim_{x \to \infty} \frac{8x^4 - 3x^2 + 1}{2x^4 + 7x - 5}x→∞lim​2x4+7x−58x4−3x2+1​

Solution Key:
  • Both numerator and denominator have degree 444 (balanced).
  • Divide by x4x^4x4:
    lim⁡x→∞8−3x2+1x42+7x3−5x4=8−0+02+0−0=82=4\lim_{x \to \infty} \frac{8 - \frac{3}{x^2} + \frac{1}{x^4}}{2 + \frac{7}{x^3} - \frac{5}{x^4}} = \frac{8 - 0 + 0}{2 + 0 - 0} = \frac{8}{2} = 4x→∞lim​2+x37​−x45​8−x23​+x41​​=2+0−08−0+0​=28​=4
  • Horizontal Asymptote: y=4y = 4y=4.
Show your steps and calculations here.

Practice Problem 13

Evaluate the rational limit and state the horizontal asymptote:

lim⁡x→∞14x+93x2−5x+2\lim_{x \to \infty} \frac{14x + 9}{3x^2 - 5x + 2}x→∞lim​3x2−5x+214x+9​

Solution Key:
  • Numerator degree is 111, denominator degree is 222 (bottom-heavy).
  • Divide by x2x^2x2:
    lim⁡x→∞14x+9x23−5x+2x2=0+03−0+0=03=0\lim_{x \to \infty} \frac{\frac{14}{x} + \frac{9}{x^2}}{3 - \frac{5}{x} + \frac{2}{x^2}} = \frac{0 + 0}{3 - 0 + 0} = \frac{0}{3} = 0x→∞lim​3−x5​+x22​x14​+x29​​=3−0+00+0​=30​=0
  • Horizontal Asymptote: y=0y = 0y=0.
Show your steps and calculations here.

Regroup 2

  • Key Reflection: When evaluating limits at infinity, why does dividing by the highest power of the denominator work universally, even for top-heavy and bottom-heavy functions?
  • Degree Taxonomy Check: Have students practice quickly identifying whether a rational function has a horizontal asymptote of y=0y = 0y=0, y=an/bmy = a_n/b_my=an​/bm​, or no HA simply by scanning the leading polynomial terms.

Skill Block 3

Worked Example A: Indeterminate Radical Form at Infinity (∞−∞\infty - \infty∞−∞)

Evaluate the limit:

lim⁡x→∞(x2+8x−x)\lim_{x \to \infty} \left( \sqrt{x^2 + 8x} - x \right)x→∞lim​(x2+8x​−x)

Worked Solution:
  1. Recognize Indeterminate Form: Direct substitution gives ∞2+∞−∞=∞−∞\sqrt{\infty^2 + \infty} - \infty = \infty - \infty∞2+∞​−∞=∞−∞ (indeterminate).
  2. Multiply and Divide by the Conjugate (x2+8x+x)(\sqrt{x^2 + 8x} + x)(x2+8x​+x):
    lim⁡x→∞(x2+8x−x)(x2+8x+x)x2+8x+x=lim⁡x→∞(x2+8x)−x2x2+8x+x=lim⁡x→∞8xx2+8x+x\lim_{x \to \infty} \frac{(\sqrt{x^2 + 8x} - x)(\sqrt{x^2 + 8x} + x)}{\sqrt{x^2 + 8x} + x} = \lim_{x \to \infty} \frac{(x^2 + 8x) - x^2}{\sqrt{x^2 + 8x} + x} = \lim_{x \to \infty} \frac{8x}{\sqrt{x^2 + 8x} + x}x→∞lim​x2+8x​+x(x2+8x​−x)(x2+8x​+x)​=x→∞lim​x2+8x​+x(x2+8x)−x2​=x→∞lim​x2+8x​+x8x​
  3. Divide Numerator and Denominator by xxx (x=x2x = \sqrt{x^2}x=x2​ for x>0x > 0x>0):
    lim⁡x→∞8xxx2+8xx2+xx=lim⁡x→∞81+8x+1\lim_{x \to \infty} \frac{\frac{8x}{x}}{\frac{\sqrt{x^2 + 8x}}{\sqrt{x^2}} + \frac{x}{x}} = \lim_{x \to \infty} \frac{8}{\sqrt{1 + \frac{8}{x}} + 1}x→∞lim​x2​x2+8x​​+xx​x8x​​=x→∞lim​1+x8​​+18​
  4. Evaluate the Limit:
    81+0+1=81+1=82=4\frac{8}{\sqrt{1 + 0} + 1} = \frac{8}{1 + 1} = \frac{8}{2} = 41+0​+18​=1+18​=28​=4

Worked Example B: Radical Limit at Negative Infinity (x→−∞x \to -\inftyx→−∞)

Evaluate the limit and state the horizontal asymptote as x→−∞x \to -\inftyx→−∞:

lim⁡x→−∞9x2+42x−5\lim_{x \to -\infty} \frac{\sqrt{9x^2 + 4}}{2x - 5}x→−∞lim​2x−59x2+4​​

Worked Solution:
  1. Identify the Sign of xxx and Denominator Dominance: As x→−∞x \to -\inftyx→−∞, xxx is negative (x<0x < 0x<0). The dominant denominator power is x1x^1x1.
  2. Apply the Negative Radical Rule (x2=−x  ⟹  x=−x2\sqrt{x^2} = -x \implies x = -\sqrt{x^2}x2​=−x⟹x=−x2​): Dividing the numerator by xxx (for x<0x < 0x<0):
    9x2+4x=9x2+4−x2=−9x2+4x2=−9+4x2\frac{\sqrt{9x^2 + 4}}{x} = \frac{\sqrt{9x^2 + 4}}{-\sqrt{x^2}} = -\sqrt{\frac{9x^2 + 4}{x^2}} = -\sqrt{9 + \frac{4}{x^2}}x9x2+4​​=−x2​9x2+4​​=−x29x2+4​​=−9+x24​​
  3. Divide Denominator by xxx:
    2x−5x=2−5x\frac{2x - 5}{x} = 2 - \frac{5}{x}x2x−5​=2−x5​
  4. Evaluate the Limit:
    lim⁡x→−∞−9+4x22−5x=−9+02−0=−32\lim_{x \to -\infty} \frac{-\sqrt{9 + \frac{4}{x^2}}}{2 - \frac{5}{x}} = \frac{-\sqrt{9 + 0}}{2 - 0} = -\frac{3}{2}x→−∞lim​2−x5​−9+x24​​​=2−0−9+0​​=−23​
    • Limit: −32-\frac{3}{2}−23​
    • Horizontal Asymptote (x→−∞x \to -\inftyx→−∞): y=−32y = -\frac{3}{2}y=−23​

Practice Problem 14: Concept & Reading Check

Open this Skill Block on the Memex to locate the information needed to answer the following:

  • (a) Under the Conceptual Framework, why is the expression ∞−∞\infty - \infty∞−∞ classified as an indeterminate form rather than evaluating to 000? How does the conjugate method resolve this ambiguity?
  • (b) In Worked Example B, why did dividing 9x2+4\sqrt{9x^2 + 4}9x2+4​ by xxx introduce an explicit negative sign (−9+4/x2-\sqrt{9 + 4/x^2}−9+4/x2​)?
  • (c) Review the linked Engineering Applications section. How do aerospace engineers utilize limits at infinity of exponential models to determine the terminal velocity in the Terminal Velocity of a Re-entry Capsule (Aerospace Engineering) model?
Solution Key:
  • (a) In calculus, infinity is not a fixed real number; subtracting two unbounded expressions leaves open whether one grows faster, both grow at the same rate, or they differ by a constant. Multiplying by A+BA+B\frac{\sqrt{A}+B}{\sqrt{A}+B}A​+BA​+B​ eliminates the radical difference in the numerator, producing a standard ratio that can be evaluated via Growth Dominance.
  • (b) For any real number, x2=∣x∣\sqrt{x^2} = |x|x2​=∣x∣. When x<0x < 0x<0, ∣x∣=−x|x| = -x∣x∣=−x, meaning x=−x2x = -\sqrt{x^2}x=−x2​. Therefore, 9x2+4x=9x2+4−x2=−9x2+4x2=−9+4x2\frac{\sqrt{9x^2+4}}{x} = \frac{\sqrt{9x^2+4}}{-\sqrt{x^2}} = -\sqrt{\frac{9x^2+4}{x^2}} = -\sqrt{9 + \frac{4}{x^2}}x9x2+4​​=−x2​9x2+4​​=−x29x2+4​​=−9+x24​​.
  • (c) In the capsule model v(t)=76e0.4t−122e0.4t+5v(t) = \frac{76e^{0.4t}-12}{2e^{0.4t}+5}v(t)=2e0.4t+576e0.4t−12​, dividing by e0.4te^{0.4t}e0.4t gives 76−12e−0.4t2+5e−0.4t\frac{76 - 12e^{-0.4t}}{2 + 5e^{-0.4t}}2+5e−0.4t76−12e−0.4t​. As t→∞t \to \inftyt→∞, e−0.4t→0e^{-0.4t} \to 0e−0.4t→0, giving the terminal velocity vterm=762=38 m/sv_{\text{term}} = \frac{76}{2} = 38\text{ m/s}vterm​=276​=38 m/s.
Show your steps and calculations here.

Practice Problem 15

Evaluate the limit analytically using conjugate rationalization:

lim⁡x→∞(4x2+12x−2x)\lim_{x \to \infty} \left( \sqrt{4x^2 + 12x} - 2x \right)x→∞lim​(4x2+12x​−2x)

Solution Key:
  1. Multiply and Divide by the Conjugate:
    lim⁡x→∞(4x2+12x−2x)(4x2+12x+2x)4x2+12x+2x=lim⁡x→∞(4x2+12x)−4x24x2+12x+2x=lim⁡x→∞12x4x2+12x+2x\lim_{x \to \infty} \frac{(\sqrt{4x^2 + 12x} - 2x)(\sqrt{4x^2 + 12x} + 2x)}{\sqrt{4x^2 + 12x} + 2x} = \lim_{x \to \infty} \frac{(4x^2 + 12x) - 4x^2}{\sqrt{4x^2 + 12x} + 2x} = \lim_{x \to \infty} \frac{12x}{\sqrt{4x^2 + 12x} + 2x}x→∞lim​4x2+12x​+2x(4x2+12x​−2x)(4x2+12x​+2x)​=x→∞lim​4x2+12x​+2x(4x2+12x)−4x2​=x→∞lim​4x2+12x​+2x12x​
  2. Divide Numerator and Denominator by xxx (x=x2x = \sqrt{x^2}x=x2​ for x>0x > 0x>0):
    lim⁡x→∞124x2+12xx2+2=lim⁡x→∞124+12x+2\lim_{x \to \infty} \frac{12}{\sqrt{\frac{4x^2 + 12x}{x^2}} + 2} = \lim_{x \to \infty} \frac{12}{\sqrt{4 + \frac{12}{x}} + 2}x→∞lim​x24x2+12x​​+212​=x→∞lim​4+x12​​+212​
  3. Evaluate the Limit:
    =124+0+2=122+2=124=3= \frac{12}{\sqrt{4 + 0} + 2} = \frac{12}{2 + 2} = \frac{12}{4} = 3=4+0​+212​=2+212​=412​=3
Show your steps and calculations here.

Practice Problem 16

Evaluate the function f(x)=6x−14x2+9f(x) = \frac{6x - 1}{\sqrt{4x^2 + 9}}f(x)=4x2+9​6x−1​ as:

  • (a) x→∞x \to \inftyx→∞
  • (b) x→−∞x \to -\inftyx→−∞
  • (c) State all horizontal asymptotes of f(x)f(x)f(x).
Solution Key:
  1. Part (a) — As x→∞x \to \inftyx→∞ (x>0  ⟹  x=x2x > 0 \implies x = \sqrt{x^2}x>0⟹x=x2​):
    lim⁡x→∞6−1x4x2+9x2=lim⁡x→∞6−1x4+9x2=6−04+0=62=3\lim_{x \to \infty} \frac{6 - \frac{1}{x}}{\sqrt{\frac{4x^2 + 9}{x^2}}} = \lim_{x \to \infty} \frac{6 - \frac{1}{x}}{\sqrt{4 + \frac{9}{x^2}}} = \frac{6 - 0}{\sqrt{4 + 0}} = \frac{6}{2} = 3x→∞lim​x24x2+9​​6−x1​​=x→∞lim​4+x29​​6−x1​​=4+0​6−0​=26​=3
  2. Part (b) — As x→−∞x \to -\inftyx→−∞ (x<0  ⟹  x=−x2x < 0 \implies x = -\sqrt{x^2}x<0⟹x=−x2​):
    lim⁡x→−∞6−1x−4x2+9x2=lim⁡x→−∞6−1x−4+9x2=6−0−4+0=6−2=−3\lim_{x \to -\infty} \frac{6 - \frac{1}{x}}{-\sqrt{\frac{4x^2 + 9}{x^2}}} = \lim_{x \to -\infty} \frac{6 - \frac{1}{x}}{-\sqrt{4 + \frac{9}{x^2}}} = \frac{6 - 0}{-\sqrt{4 + 0}} = \frac{6}{-2} = -3x→−∞lim​−x24x2+9​​6−x1​​=x→−∞lim​−4+x29​​6−x1​​=−4+0​6−0​=−26​=−3
  3. Part (c) — Horizontal Asymptotes:
    • Right-side asymptote: y=3y = 3y=3
    • Left-side asymptote: y=−3y = -3y=−3
Show your steps and calculations here.

Practice Problem 17

For the saturation model g(x)=243+5e−0.2xg(x) = \frac{24}{3 + 5e^{-0.2x}}g(x)=3+5e−0.2x24​:

  • (a) Evaluate lim⁡x→∞g(x)\lim_{x \to \infty} g(x)limx→∞​g(x).
  • (b) Evaluate lim⁡x→−∞g(x)\lim_{x \to -\infty} g(x)limx→−∞​g(x).
  • (c) State all horizontal asymptotes of g(x)g(x)g(x).
Solution Key:
  1. Part (a) — As x→∞x \to \inftyx→∞: Since lim⁡x→∞e−0.2x=0\lim_{x \to \infty} e^{-0.2x} = 0limx→∞​e−0.2x=0:
    lim⁡x→∞243+5e−0.2x=243+5(0)=243=8\lim_{x \to \infty} \frac{24}{3 + 5e^{-0.2x}} = \frac{24}{3 + 5(0)} = \frac{24}{3} = 8x→∞lim​3+5e−0.2x24​=3+5(0)24​=324​=8
  2. Part (b) — As x→−∞x \to -\inftyx→−∞: As x→−∞x \to -\inftyx→−∞, −0.2x→+∞-0.2x \to +\infty−0.2x→+∞, which means e−0.2x→∞e^{-0.2x} \to \inftye−0.2x→∞.
    lim⁡x→−∞243+5e−0.2x→24∞=0\lim_{x \to -\infty} \frac{24}{3 + 5e^{-0.2x}} \to \frac{24}{\infty} = 0x→−∞lim​3+5e−0.2x24​→∞24​=0
  3. Part (c) — Horizontal Asymptotes:
    • As x→∞x \to \inftyx→∞: y=8y = 8y=8
    • As x→−∞x \to -\inftyx→−∞: y=0y = 0y=0
Show your steps and calculations here.

Regroup 3

  • Key Reflection: Why does the function f(x)=6x−14x2+9f(x) = \frac{6x-1}{\sqrt{4x^2+9}}f(x)=4x2+9​6x−1​ possess two distinct horizontal asymptotes (y=3y = 3y=3 and y=−3y = -3y=−3), whereas standard polynomial rational functions possess at most one?
  • Conjugate Rationalization Strategy: Emphasize that whenever students encounter ∞−∞\infty - \infty∞−∞ with square roots, conjugate multiplication converts subtraction into addition in the denominator, eliminating ambiguity.

Synthesis Wrap-up

  • Mastery Summary:
    • Vertical Asymptotes: Occur at x=ax = ax=a when lim⁡x→a±f(x)=±∞\lim_{x \to a^\pm} f(x) = \pm\inftylimx→a±​f(x)=±∞ (c0\frac{c}{0}0c​ form). Use directional signed zero analysis (0+0^+0+ vs 0−0^-0−).
    • Horizontal Asymptotes: Occur at y=Ly = Ly=L when lim⁡x→±∞f(x)=L\lim_{x \to \pm\infty} f(x) = Llimx→±∞​f(x)=L. Filter using Growth Dominance (bottom-heavy →0\to 0→0, balanced →an/bm\to a_n/b_m→an​/bm​, top-heavy →±∞\to \pm\infty→±∞).
    • Radicals & Exponentials: Use conjugates for ∞−∞\infty - \infty∞−∞, remember x2=−x\sqrt{x^2} = -xx2​=−x for x<0x < 0x<0, and evaluate exponential decays to steady-state ceilings.
  • Knewton Alta Bridge: Completes the foundation for homework assignment Infinite Limits and Asymptotes and prepares students for Session 05: Special Limits and Continuity.